Isnt setprecision not supposed to change the value stored in variable?

I was of the opinion that setprecision doesnt change the value in variable itself. Also, when you attach setprecision to cout, it sticks with i开发者_如何学Got only once. However, when I run code to verify, it doesnt work.

Consider the following code snippet:

int main()
{
    double x = 9.87654321;
    cout << setprecision(3) << fixed << x <<endl;    //Returns 9.877 as it should
    cout << x << endl;                               //Returns truncated value 9.877 again though it shouldnt.

    return 0;
}   

Interesting part is, if we replace cout << x << endl; by a line setting precision to say 7, then it DOES display the correct value. Can anyone please explain this phenomenon?

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You don't reset the precision to the original value so it's just using 3 as the precision value for both output operations.

If you want to restore the original precision then you need to save it. The initial value for standard ostreams is 6 which may not be accurate enough for many purposes.

int main()
{
    double x = 9.87654321;

    size_t save_prec = cout.precision();
    cout << setprecision(3) << fixed << x <<endl;
    cout.precision(save_prec);

    cout << x << endl;

    return 0;
}

---

It sets the precision of the output stream, not x, meaning after the call to setprecision(3), cout outputs all numbers with precision of 3.

http://www.cplusplus.com/reference/iostream/manipulators/setprecision/

Try this, which will show that x has not changed.

int main() 
{     
  double x = 9.87654321;     
  cout << setprecision(3) << fixed << x <<endl;    //Returns 9.877 as it should     
  cerr << x << endl;                               //Returns 9.87654 to stderr  
  return 0; 
} 

---

Other answers have indicated the problem. One possible solution is to use the Boost I/O Stream state saver library to do the work of saving the I/O stream formatting state and restoring it at the appropriate time.